Derivation of this function..?

Hello, TrumBeta!

Welcome aboard!

I must assume you know the Quotient Rule: . $\left(\frac{u}{v}\right)' \;=\;\frac{v\!\cdot\! u' - u\!\cdot\! v'}{v^2}$

Quote:

$y \;=\;\frac{1+x\sqrt{x}}{1 - x\sqrt{x}}$

We have: . $y \;=\;\frac{1 + x^{\frac{3}{2}}}{1 - x^{\frac{3}{2}}}$

Then: . $y' \;=\;\frac{\left(1-x^{\frac{3}{2}}\right) \left(\frac{3}{2}x^{\frac{1}{2}}\right) - \left(1+x^{\frac{3}{2}}\right)\left(-\frac{3}{2}x^{\frac{1}{2}}\right)} {\left(1 - x^{\frac{3}{2}}\right)^2}$

. . . . . . . $= \;\frac{\frac{3}{2}x^{\frac{1}{2}} - \frac{3}{2}x^2 + \frac{3}{2}x^{\frac{1}{2}} + \frac{3}{2}x^2}{\left(1-x^{\frac{3}{2}}\right)^2}$

. . . . . . . $= \;\frac{3x^{\frac{1}{2}}}{\left(1-x^{\frac{3}{2}}\right)^2}$

Quote:

$y \;=\;\frac{e^x + e^{-x}}{e^x - e^{-x}}$

$y' \;=\;\frac{(e^x-e^{-x})(e^x-e^{-x}) - (e^x + e^{-x})(e^x+e^{-x})} {(e^x-e^{-x})^2}$

. . $=\; \frac{\left(e^{2x} - 1 - 1 + e^{-2x}\right) - \left(e^{2x} + 1 + 1 + e^{-2x}\right)}{\left(e^x-e^{-x}\right)^2}<br />$

. . $=\; \frac{e^{2x} - 2 + e^{-2x} - e^{2x} - 2 - e^{-2x}}{(e^x - e^{-x})^2}$

. . $=\; \frac{-4}{(e^x-e^{-x})^2}$